If the curves y^2=6x and 9x^2+by^2=16 interse | vedclass

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(A) Given curves are $y^2=6x$ $(i)$ and $9x^2+by^2=16$ (ii).Differentiating $(i)$ with respect to $x$: $2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx}

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$\frac{9}{2}$

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(A) Given curves are $y^2=6x$ $(i)$ and $9x^2+by^2=16$ (ii).Differentiating $(i)$ with respect to $x$: $2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y}$.Differentiating (ii) with respect to $x$: $18x + 2by \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{9x}{by}$.Since the curves intersect at right angles,the product of their slopes at the point of intersection $(x, y)$ must be $-1$.$\left(\frac{3}{y}ight) 	imes \left(-\frac{9x}{by}ight) = -1$.$\Rightarrow \frac{27x}{by^2} = 1 \Rightarrow by^2 = 27x$.Substitute $y^2=6x$ from $(i)$ into this equation: $b(6x) = 27x$.Assuming $x 
eq 0$,we get $6b = 27 \Rightarrow b = \frac{27}{6} = \frac{9}{2}$.

If the curves $y^2=6x$ and $9x^2+by^2=16$ intersect each other at right angles,then the value of $b$ is

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